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Question

# If the endpoints of the diameter is P(3,1) and Q(h,k) of the circle 2x2+2y2−2x−7y−7=0, then which of the following is/are correct?

A
Q=(1,2)
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B
Q=(2,52)
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C
radius of circle is 674 units
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D
radius of circle is 1094 units
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Solution

## The correct options are B Q=(−2,52) D radius of circle is √1094 unitsGiven: 2x2+2y2−2x−7y−7=0 ⇒x2+y2−x−7y2−72=0g=−12,f=−74,c=−72 C=(−g,−f)=(12,74)r=√(−12)2+(−74)2+72⇒r=√1094 Endpoint of the diameter are P(3,1) and Q(h,k) Centre is the mid point of the two end points of diameter, so h+32=12,k+12=74⇒h=−2,k=52∴Q=(−2,52)

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